Let h(x)=f(x)-g(x), and Lagrange mean value theorem exists: c 1∈(a, x0), c2∈(x0, b).

Let h' (c1) = (h (x0)-h (a))/(x0-a) = h (x0)/(x0-a), and h' (C2) = (h (b)-h (x0))/(b

∵h(x0)>0，x0-a & gt； 0，b-x0 & gt； 0,∴h'(c 1)>； 0 & gth'(C2) is derived from Lagrange mean value theorem.

ξ∈(c 1, c2) exists, so h'' (ξ) = [h' (C2)-h' (c1)]/(C2-c1), ∫. x0 & gtc 1，h '(C2)& lt； h'(c 1)

∴h''(ξ)<； 0, namely ξ∈(a, b), f'' (ξ) < g''(ξ)